Htykuut

Prove that $$int_0^frac{npi}{4} left(frac{1}{3sin^4(x) + 3cos^4(x) -1}right) dx = frac{n pi}{4} $$
is true for all integers $n$.

Here I've looked at the case where $n=1$ as $t=tan(x)$ is injective in that domain. Then I could perhaps try to use some argument about periodicity and symmetry to extend it to all the integers.

$$
= int_0^frac{pi}{4} frac{sec^4(x)}{3tan^4(x)+3-sec^4(x)} text{d}x
= frac12 int_0^1 frac{t^2+1}{t^4-t^2+1} text{d}t
$$

Which looks a little tough but reminiscent of the well-known integral:

$$ int_0^infty frac{1}{x^4 + 2x^2cos(2a) + 1} text{d}x = frac12int_0^infty frac{x^2+1}{x^4 + 2x^2cos(2a) + 1} text{d}x = frac{pi}{4cos(a)} $$

Where we choose $a=fracpi3$. But I can't seem to adjust the bounds to get an answer. How would you evaluate this integral?

For $x in (0,pi/4)$,
the integral $$frac12 int_0^1 frac{t^2+1}{t^4-t^2+1} text{d}t$$ can also be written as $$frac12 int_0^1 frac{1+1/t^2}{(t-1/t)^2+1} text{d}t$$
Substituting $u=t-frac1t$, the integral becomes $$frac12 int_{-infty}^0 frac{1}{u^2+1} text{d}u$$ which is equal to $$frac12[arctan(u)]_{-infty}^0=pi/4$$
Similarly for $x in (pi/4,pi/2)$, the integral becomes$$frac12 int_1^infty frac{1+1/t^2}{(t-1/t)^2+1} text{d}t$$ which is equal to$$frac12 int_0^{infty} frac{1}{v^2+1} text{d}v$$ or$$frac12[arctan(u)]_0^{infty}=pi/4$$
And since the integrand has a period of $pi/2$ because $$sin^4x+cos^4x=1-frac12sin^22x$$
one can easily contemplate the proof by adding the areas of the curve on and on.

$$int_0^{pi/4}frac{1}{3sin^4(x) + 3cos^4(x) -1} dx = frac{pi}{4}$$
can be shown by using partial fraction in OP's substituted integral.
One can similarly show $$int_{pi/4}^{pi/2}frac{1}{3sin^4(x) + 3cos^4(x) -1} dx = frac{pi}{4}$$
Observe that the integrand has period $fracpi2$, the problem can be easily done by using induction.

$$3(sin^4x+cos^4x)-1=3((sin^2x+cos^2x)^2-2sin^2xcos^2x)-1=2-frac32sin^22x=frac{5+3cos4x}4$$

and your integral is also

$$int_0^{npi}frac{dx}{5+3cos x}=nint_0^{pi}frac{dx}{5+3cos x}$$ as the cosine is an even function.

Using the Weierstrass substitution, the last integral is shown to be $dfracpi4$.

$$int_0^pifrac{dx}{5+3cos4x}=int_0^inftyfrac{2,dt}{(1+t^2)left(5+3dfrac{1-t^2}{1+t^2}right)}=int_0^inftyfrac{dt}{4+t^2}.$$

Note that the integrand function $f(x)=frac{1}{3sin^4(x) + 3cos^4(x) -1}$ is even with period $pi/2$, so
$$int_0^frac{npi}{4} f(x) dx=
frac{1}{2}int_{-frac{npi}{4}}^frac{npi}{4} f(x) dx=
frac{1}{2}int_{0}^frac{npi}{2} f(x) dx
=frac{n}{2}int_{0}^frac{pi}{2} f(x) dx.$$
Now, following your approach, we have that
$$frac{t^2+1}{t^4-t^2+1} =
frac{2}{1+(2t+sqrt{3})^2} +frac{2}{1+(2t-sqrt{3})^2}.$$
Therefore
$$intfrac{t^2+1}{t^4-t^2+1},dt=arctan(2t+sqrt{3})+arctan(2t-sqrt{3})+c$$
Hence, for $t=tan(x)$,
$$int_0^frac{pi}{2} f(x) dx
= frac12 int_0^{infty} frac{t^2+1}{t^4-t^2+1} dt=frac{1}{2}left[arctan(2t+sqrt{3})+arctan(2t-sqrt{3})right]_0^{+infty}=frac{pi}{2}.$$